The Shape and Action of a Nilpotent Jordan Block
A nilpotent Jordan block is a basic matrix for studying nilpotent linear maps. Its entries are sparse, but the positions of its ones encode a sequence: each application of the map moves a standard basis vector one position down the sequence, until it reaches zero. Following this action makes the powers and kernels of the block accessible without multiplying large matrices blindly.
Throughout, work over \(\mathbb{R}\), and let \(k\) be a positive integer. The standard basis of \(\mathbb{R}^k\) is \(\mathbf{e}_1,\dots,\mathbf{e}_k\), where \(\mathbf{e}_j\) has coordinate \(1\) in position \(j\) and \(0\) elsewhere. We use the convention that the ones in a Jordan block lie immediately above the diagonal.
For \(k\ge 2\), its general appearance is
Matrix multiplication shows directly what the associated linear map does to the standard basis. The \(j\)th column of a matrix is its product with \(\mathbf{e}_j\). The first column of \(\mathbf{J}_k(0)\) is zero, while column \(j\), for \(2\le j\le k\), has its \(1\) in row \(j-1\). Therefore \(\mathbf{J}_k(0)\mathbf{e}_1=\mathbf{0}\) and \(\mathbf{J}_k(0)\mathbf{e}_j=\mathbf{e}_{j-1}\) for \(j\ge2\).
The indexing matters: the block sends \(\mathbf{e}_k\) to \(\mathbf{e}_{k-1}\), not to \(\mathbf{e}_{k+1}\). Repeated application moves left through the subscripts: \(\mathbf{e}_k\) goes to \(\mathbf{e}_{k-1}\), then \(\mathbf{e}_{k-2}\), and eventually to \(\mathbf{0}\).
Deriving the Formula for Powers
For a nonnegative integer \(m\), \(\mathbf{J}_k(0)^m\) means the product of \(m\) copies of the matrix, with the convention \(\mathbf{J}_k(0)^0=\mathbf{I}_k\). The action on each basis vector gives a compact formula for every power. A basis vector is shifted down \(m\) positions if it has at least \(m\) positions to move; otherwise its image is zero.
Proof. We prove the action formula by induction on \(m\). When \(m=0\), the power is \(\mathbf{I}_k\), so \(\mathbf{J}_k(0)^0\mathbf{e}_j=\mathbf{e}_j\) for every \(j\); this is the first case of the formula because \(j>0\). Now suppose the formula holds for some \(m\ge0\). If \(j\le m\), then \(\mathbf{J}_k(0)^m\mathbf{e}_j=\mathbf{0}\), so applying \(\mathbf{J}_k(0)\) again still gives \(\mathbf{0}\). If \(j=m+1\), the induction hypothesis gives \(\mathbf{e}_1\), and \(\mathbf{J}_k(0)\mathbf{e}_1=\mathbf{0}\). If \(j>m+1\), the induction hypothesis gives \(\mathbf{e}_{j-m}\), whose subscript is at least \(2\); hence one more application gives \(\mathbf{e}_{j-m-1}\). These cases establish the formula for \(m+1\). The columns of a matrix are its products with the standard basis vectors, so the formula puts a \(1\) in column \(i+m\), row \(i\), exactly when \(i+m\le k\). If \(m\ge k\), every \(j\) satisfies \(j\le m\), so every column is zero. \(\square\)
This theorem also gives a useful description of the kernel. The kernel of a matrix \(\mathbf{A}\) is the set \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). Write \(\mathbf{x}=x_1\mathbf{e}_1+\cdots+x_k\mathbf{e}_k\). For \(0\le m<k\), the action formula gives \(\mathbf{J}_k(0)^m\mathbf{x}=x_{m+1}\mathbf{e}_1+\cdots+x_k\mathbf{e}_{k-m}\). These are distinct standard basis vectors, so the result is zero exactly when \(x_{m+1},\dots,x_k\) are all zero. Thus the kernel consists of the vectors supported in the first \(m\) coordinates. For \(m=0\), this means only the zero vector; for \(m\ge k\), the whole space is the kernel because the power is the zero matrix.
Worked Powers and Kernels
Worked Example: Powers of a Block of Size Three
Compute the powers of \(\mathbf{J}_3(0)\) and find the kernel of its first power. The block is
Multiplying \(\mathbf{J}_3(0)\) by itself gives
For instance, the entry in row \(1\), column \(3\) is \(0\cdot0+1\cdot1+0\cdot0=1\); the other entries follow by the same row-by-column multiplication. Multiplying once more, the only possible nonzero entry in row \(1\) of \(\mathbf{J}_3(0)^2\) lies in column \(3\), while column \(3\) of \(\mathbf{J}_3(0)\) is \((0,1,0)^{\top}\). Their dot product is \(0\). All other rows of \(\mathbf{J}_3(0)^2\) are zero, so \(\mathbf{J}_3(0)^3=\mathbf{O}\).
For a general vector \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\),
This is zero exactly when \(x_2=0\) and \(x_3=0\); \(x_1\) is unrestricted. Therefore \(\operatorname{Nul}\mathbf{J}_3(0)=\{(x_1,0,0)^{\top}:x_1\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1\}\).
Check: \(\mathbf{J}_3(0)(x_1,0,0)^{\top}=(0,0,0)^{\top}\), as required. Also, \(\mathbf{J}_3(0)^2\mathbf{e}_3=\mathbf{e}_1\) and \(\mathbf{J}_3(0)^2\mathbf{e}_1=\mathbf{J}_3(0)^2\mathbf{e}_2=\mathbf{0}\), agreeing with the displayed square.
Worked Example: The Square of a Block of Size Four
Compute \(\mathbf{J}_4(0)^2\) and its kernel. The block is
Each row of the first factor selects one row of the second factor: row \(1\) selects row \(2\), row \(2\) selects row \(3\), and rows \(3\) and \(4\) give zero. Thus
In particular, the entry in row \(1\), column \(3\) is the dot product of row \(1\), \((0,1,0,0)\), with column \(3\), \((0,1,0,0)^{\top}\), of the second factor:
For \(\mathbf{x}=(x_1,x_2,x_3,x_4)^{\top}\), the displayed square gives \(\mathbf{J}_4(0)^2\mathbf{x}=(x_3,x_4,0,0)^{\top}\). This is zero precisely when \(x_3=x_4=0\). Hence \(\operatorname{Nul}\mathbf{J}_4(0)^2=\{(x_1,x_2,0,0)^{\top}:x_1,x_2\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\).
Check: multiplying the displayed square by \(\mathbf{x}\) yields \((x_3,x_4,0,0)^{\top}\); if \(\mathbf{x}\) is in the stated kernel, this product is \(\mathbf{0}\). The columns of the square are \(\mathbf{0},\mathbf{0},\mathbf{e}_1,\mathbf{e}_2\), which also matches the action formula for the second power.
Worked Example: A Third Power of a Block of Size Five
Compute \(\mathbf{J}_5(0)^3\) and determine its kernel. The action formula says that the third power sends \(\mathbf{e}_4\) to \(\mathbf{e}_1\), sends \(\mathbf{e}_5\) to \(\mathbf{e}_2\), and sends \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\) to zero. Therefore its columns give
As a direct product check, multiplying this matrix by \(\mathbf{J}_5(0)\) sends \(\mathbf{e}_5\) to \(\mathbf{e}_1\) and sends every other standard basis vector to zero. This is the action of the fourth power, so
Multiplying this matrix by \(\mathbf{J}_5(0)\) gives the zero matrix: its only nonzero row has a \(1\) in column \(5\), and row \(5\) of \(\mathbf{J}_5(0)\) is zero. Hence \(\mathbf{J}_5(0)^5=\mathbf{O}\), also as predicted by the formula.
For \(\mathbf{x}=(x_1,x_2,x_3,x_4,x_5)^{\top}\), the third power gives \((x_4,x_5,0,0,0)^{\top}\). Its kernel is therefore \(\{(x_1,x_2,x_3,0,0)^{\top}:x_1,x_2,x_3\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}\).
Check: applying the displayed matrix to any vector in this span gives \((0,0,0,0,0)^{\top}\). Conversely, if \(\mathbf{J}_5(0)^3\mathbf{x}=\mathbf{0}\), the first two coordinates of the product force \(x_4=x_5=0\), so every vector in the kernel has the stated form.
Reading a Kernel from the Shift
The examples exhibit the same mechanism. The \(m\)th power kills the first \(m\) standard basis vectors because they do not have \(m\) positions left to shift. Any later basis vector survives and is sent to a different standard basis vector. Different surviving basis vectors cannot cancel one another: their images occupy different coordinates. That is why solving for the kernel reduces to setting the corresponding coordinates of the input equal to zero.
For a single block, this coordinate description is often more informative than expanding a general matrix product. To compute \(\operatorname{Nul}\mathbf{J}_k(0)^m\), first identify which basis vectors survive, then write the image of a general vector in those coordinates. The surviving input coordinates must vanish; the others may be chosen freely. This procedure handles the endpoints as well: the zeroth power is the identity, so its kernel contains only \(\mathbf{0}\), and every power with \(m\ge k\) is zero, so its kernel is all of \(\mathbb{R}^k\).
Common Pitfall: Reversing the Shift
A frequent error is to read the block as sending \(\mathbf{e}_j\) to \(\mathbf{e}_{j+1}\). That would move basis vectors in the wrong direction and would contradict the columns of the displayed matrix. For example, in \(\mathbf{J}_4(0)\), column \(3\) is \((0,1,0,0)^{\top}=\mathbf{e}_2\), so \(\mathbf{J}_4(0)\mathbf{e}_3=\mathbf{e}_2\), not \(\mathbf{e}_4\).
Another concrete error can occur during multiplication: using row \(1\) of \(\mathbf{J}_4(0)\) and column \(3\) of the second factor, one might mistakenly write \(0\cdot0+1\cdot0+0\cdot0+0\cdot0=0\). The second entry of column \(3\) is \(1\), not \(0\). The corrected calculation is \(0\cdot0+1\cdot1+0\cdot0+0\cdot0=1\), which is exactly the row \(1\), column \(3\) entry of \(\mathbf{J}_4(0)^2\). Checking the row and column explicitly prevents both the arithmetic error and an incorrect power matrix.
Check Your Understanding
Use the basis-vector action and verify each result by multiplying a general coordinate vector or checking the relevant columns.
- Write \(\mathbf{J}_4(0)\mathbf{e}_1,\dots,\mathbf{J}_4(0)\mathbf{e}_4\).
- Compute \(\mathbf{J}_3(0)^2\) and \(\mathbf{J}_3(0)^3\).
- Find \(\operatorname{Nul}\mathbf{J}_4(0)^3\) by applying the power to \((x_1,x_2,x_3,x_4)^{\top}\).
- For \(\mathbf{J}_5(0)^2\), state which standard basis vectors map to zero and which survive.
- Explain why the nonzero columns of \(\mathbf{J}_k(0)^m\), when \(m<k\), cannot cancel in a linear combination of those columns.
